SEQUENCE October 20, 2008

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1 ON β-elements IN THE ADAMS-NOVIKOV SPECTRAL SEQUENCE October 0, 008 HIROFUMI NAKAI AND DOUGLAS C RAVENEL Dedicated to Professor Takao Matumoto on his sixtieth birthday The second author acknowledges support from NSF Grants DMS and DMS Abstract In this paper we detect invariants in the comodule consisting of β-elements over the Hopf algebroid (A(m +, G(m + defined in[rav0], and we show that some related Ext groups vanish below a certain dimension The result obtained here will be extensively used in [NR] to extend the range of our knowledge for π (T (m obtained in[rav0] Contents Introduction The construction of B m+ 5 3 Basic methods for finding comodule primitives primitives in B m+ 5 -primitives in B m+ 3 6 j-primitives in B m+ for j > 6 7 Higher Ext groups for j = Appendix A Some results on binomial coefficients Appendix B Quillen operations on β-elements 5 References 7 Introduction In this paper we describe some tools needed in the method of infinite descent, which is an approach to finding the E -term of the Adams-Novikov spectral sequence converging to the stable homotopy groups of spheres It is the subject of [Rav86, Chapter 7], [Rav04, Chapter 7] and [Rav0] We begin by reviewing some notation Fix a prime p Recall the Brown-Peterson spectrum BP Its homotopy groups and those of BP BP are known to be Date: October 0, Mathematics Subject Classification 55Q99 Key words and phrases Homotopy groups of spheres, Adams-Novikov spectral sequence

2 HIROFUMI NAKAI AND DOUGLAS C RAVENEL polynomial algebras π (BP = Z (p [v, v ] and BP (BP = BP [t, t ] In [Rav86, Chapter 6] the second author constructed intermediate spectra S 0 (p = T (0 T ( T ( T (3 BP with T (m is equivalent to BP below the dimension of v m+ This range of dimensions grows exponentially with m T (m is a summand of p-localization of the Thom spectrum of the stable vector bundle induced by the map ΩSU(p m ωsu = BU In [Rav0] we constructed truncated versions T (m (j for j 0 with T (m = T (m (0 T (m ( T (m ( T (m + These spectra satisfy BP (T (m = π (BP [t,, t m ] and BP (T (m (j = BP (T (m { t l m+ : 0 l < p j Thus T (m (j has p j cells, each of which is a copy of T (m For each m 0 we define a Hopf algebroid Γ(m + = (BP, BP (BP /(t, t,, t m = BP [t m+, t m+, ] with structure maps inherited from BP (BP, which is Γ( by definition Let A = BP, A(m = Z (p [v,, v m ] and G(m + = A(m + [t m+ ] with t m+ primitive Then there is a Hopf algebroid extension ( (A(m +, G(m + (A, Γ(m + (A, Γ(m + In order to avoid excessive subscripts, we will use the notation v i = v m+i, and t i = t m+i We will use the usual notation without hats when m = 0 We will use the notation v i = v m+i, t i = t m+i, βi/e,e 0 = vi p e0 v e and β i/e = vi piv e We will also use the notations β i/e = β i/e, and β i/e = β i/e, for short We will use the usual notation without hats when m = 0 Given a Hopf algebroid (B, Γ and a Γ-comodule M, we will abbreviate Ext Γ (B, M by Ext Γ (M and Ext Γ (B by Ext Γ With this in mind, there are change-of-rings isomorphisms Ext BP (BP (BP (T (m = Ext Γ(m+ and Ext BP (BP (BP (T (m (j = Ext Γ(m+ ( where T m (j T (j m = A { t l : 0 l < p j

3 ON β-elements IN THE ADAMS-NOVIKOV SPECTRAL SEQUENCE 3 Very briefly, the method of infinite descent involves determining the groups ( Ext Γ(m+ T (j ( and π T (m(j by downward induction on m and j To begin with, we know that m Ext 0 Γ(m+ ( A { t l m+ : 0 l < p j = A(m { v l : 0 l < p j To proceed further, we make use of a short exact sequence of Γ(m + -comodules ( 0 BP ι 0 D 0 m+ ρ 0 E m+ 0, where D 0 m+ is weak injective (meaning that its higher Ext groups vanish with ι 0 inducing an isomorphism in Ext 0 It has the form with D 0 m+ = A(m[ λ, λ, ] Q BP λ i = p v i + Thus we have an explicit description of Em+, which is a certain subcomodule of the chromatic module N = BP /(p It follows that the connecting homomorphism δ 0 associated with ( is an isomorphism Ext s Γ(m+(Em+ = Ext s+ Γ(m+ and more generally Ext s Γ(m+(E m+ T (j m = Ext s+ (j Γ(m+ (T m for each s 0 The determination of this group for s = 0 will be the subject of [Nak] In this paper we will limit our attention to the case s > 0 Unfortunately there is no way to embed E m+ in a weak injective comodule in a way that induces an isomorphism in Ext 0 as in ( (This is explained in [NR, Remark74] Instead we will study the Cartan-Eilenberg spectral sequence for Ext Γ(m+ (E m+ T (j m associated with the extension ( Its E -term is (3 Ẽ s,t (j (T m = Ext s G(m+(Ext t Γ(m+(T m (j Em+ = Ext s G(m+(T (j m Ext t Γ(m+(E m+ where T (j m = A(m + { t l : 0 l < p j and differentials d r : Ẽs,t Ẽs+r,t r+ Note that T m (j = A A(m+ T (j m We use the tilde to distinguish this spectral sequence from the resolution spectral sequence We did not use this notation in [Rav0] The short exact sequence of Γ(m + -comodules ( is also a one of Γ(m + - comodules, and D 0 m+ is also weak injective over Γ(m + (this was proved in [Rav0, Lemma ], but this time the map ι 0 does not induce an isomorphism in Ext 0 However, the connecting homomorphism δ 0 : Ext t Γ(m+(Em+ T m (j Ext t+ (j Γ(m+ (T m

4 4 HIROFUMI NAKAI AND DOUGLAS C RAVENEL is an isomorphim of G(m + -comdules for t > 0 Note that over Γ(m +, T m (j is a direct sum of p j suspended copies of A, so the isomorphism above is the tensor product with T (j m with δ 0 : Ext t Γ(m+(E m+ Ext t+ Γ(m+ We will abbreviate the group on the right by Um+ t+ Its structure up to dimension (p + p v was determined in [NR, Theorem 70] It is p-torsion for all t 0 and v -torsion for t > 0 Moreover, it is shown that each Um+ t for t is a certain suspension of Um+ below dimension p v 3 Let E m+ = Ext 0 Γ(m+(Em+ For j = 0, the Cartan-Eilenberg E -term of (3 is { Ẽ s,t (0 Ext s (T m = G(m+(E m+ for t = 0 for t Ext s G(m+(U t+ m+ While it is impossible to embed the Γ(m+-comodule Em+ into a weak injective by a map inducing an isomorphism in Ext 0, it is possible to do this for the G(m+- comodule E m+ In Theorem 4 below we will show that there is a pullback diagram of G(m + -comodules (4 0 E m+ ι W m+ ρ B m+ 0 0 E m+ v E m+ E m+/(v 0 where W m+ is weak injective, ι induces an isomorphism in Ext 0, and B m+ is the A(m + -submodule of E m+/(v generated by { v i ipv i : i > 0 The object of this paper is to study B m+ and related Ext groups Since the ith element above is β i/i, the elements of B m+ are the beta elements of the title In [NR] we construct a variant of the Cartan-Eilenberg spectral sequence converging to Ext Γ(m+ (T (j m Its Ẽ-term has the following chart: t = 0 Ext 0 (U 3 Ext (U 3 Ext (U 3 t = 0 Ext 0 (U Ext (U Ext (U t = 0 Ext 0 (D Ext 0 (W Ext 0 (B Ext (B s = 0 s = s = s = 3 where all Ext groups are over G(m+ and the tensor product signs and subscripts (equal to m + on U t+, D 0, W and B have been omitted to save space

5 ON β-elements IN THE ADAMS-NOVIKOV SPECTRAL SEQUENCE 5 Tensoring (4 with T (j m, we also have the following diagram: t = 0 Ext 0 (T (j m U 3 Ext (T (j m U 3 Ext (T (j m U 3 (5 t = 0 Ext 0 (T (j m U Ext (T (j m U Ext (T (j m U t = 0 Ext 0 (T (j m D Ext 0 (T (j m W Ext 0 (T (j m B Ext (T (j m B s = 0 s = s = s = 3 The construction of B m+ will be given in After introducing our basic methodology in 3, we determine the groups Ext 0 (T (j m B m+ for the cases j = 0, j = and j > in the next three sections Here T (j m = A(m + { t l m+ : 0 l < p j In 7 we determine the higher Ext groups for j = in a range of dimensions Our calculations require some results about binomial coefficients and Quillen operations that are collected in Appendices A and B respectively The construction of B m+ Proposition A 4-term exact sequence of G(m + -comodules The short exact sequence ( gives a 4-term exact sequence A(m + 0 U 0 m+ ι 0 A(m[p v ] ρ 0 E m+ δ 0 U m+ 0 Let V m+ = A(m[p v ]/A(m + { v i = A(m + p i : i > 0 BP /(p There is a short exact sequence of G(m + -comodules 0 V m+ E m+ U m+ 0 which is not split Proof The comodule D 0 m+ was described explicitly in [Rav0, Theorem 39] It has the form D 0 m+ = A(m[ λ, ] p BP

6 6 HIROFUMI NAKAI AND DOUGLAS C RAVENEL with and λ i = v for i = p v p + v v pω p + (pp v v p p p+ for i = v i p + for i > λ + t for i = ( p η R ( λ i = λ + t + (p p v p λ p j t j for i = j 0<j<p λ i + t i + for i > It follows that Ext 0 Γ(m+(D 0 m+ = A(m[ λ ] as claimed In order to understand the relation between E m+ and U m+, consider the following diagram of Γ(m + -comodules with exact rows 0 BP D 0 m+ E m+ 0 0 BP p BP BP /(p 0 0 BP D 0 m+ E m+ The vertical maps are monomorphisms, and there is no obvious map either way between D 0 m+ and D 0 m+ The description of the U m+ = Ext Γ(m+ above is in terms of the connecting homomorphism for the bottom row The element v i pi E m+ is invariant and maps to the similarly named element in Um+ To describe its image in terms of the cobar complex, we pull it back to v /pi i Dm+ 0 and compute its coboundary, which is d ( v /pi i = ( ( v + p t i v i /pi = v i t + However, the element v i /pi is not present in E m+ To see this, consider the case i = In p BP we have v p = λ v v pω p + ( pp v v p p p+ = λ λ v pω p + ( pp v λp p / D 0 m+ = A(m[ λ, λ, ] Instead of v /p, consider the element λ itself Its image in E m+ is invariant, so it defines a nontrivial element in E m+ The computation of the image of (p λ i /pi under the connecting homomorphism gives the same answer as before 0

7 ON β-elements IN THE ADAMS-NOVIKOV SPECTRAL SEQUENCE 7 The right unit formula above implies that the short exact sequence does not split Definition Let M be a graded torsion G(m + -comodule of finite type, and let M i have order p ai Then the Poincaré series for M is defined by (3 g(m = a i t i Given two such power series f (t and f (t, the inequality f (t f (t means that each coefficient of f (t is dominated by the corresponding one in f (t Theorem 4 Construction of B m+ A(m + -module generated by the elements Let B m+ E m+/(v be the sub- β i/i = vi ipv i for all i > 0 It is a G(m + -subcomodule whose Poincaré series is where g(b m+ = g m+ (t k 0 y = t v, x = t v, x pk+ ( y pk ( x pk+ ( x pk, x i = t vi for i > and g m+ (t = t vi i m+ Let W m+ be the pullback in the diagram (4 Then W m+ is a weak injective with Ext 0 G(m+(W m+ = Ext 0 G(m+(E m+, ie, the map E m+ W m+ induces an isomorphism in Ext 0 Proof To show that B m+ is a G(m + -subcomodule, note that η R ( v v + v t p vpω t mod p so η R ( v i = ( v + v t p vpω t p mod pi and η R ( β i/i B m+ G(m + so B m+ is a G(m + -comodule For the Poincaré series, let F k B m+ B m+ denote the submodule of exponent p k with F 0 B m+ = φ Then the Poincaré series of F k B m+ /F k B m+ = A(m + /I { βip k /ip k,p k : i > 0

8 w 8 HIROFUMI NAKAI AND DOUGLAS C RAVENEL is g (F k B m+ /F k B m+ = g(a(m + /I yipk ipk x y i>0 (x ipk (x p y ipk = g m+ (t i>0 = g m+ (t (x ipk x ipk i>0 ( x pk k = g m+ (t xp x pk x pk Summing these for all positive k gives the desired formula To show Ext 0 G(m+(W m+ is as claimed it is enough to show that the connecting homomorphism Ext 0 G(m+(B m+ Ext G(m+(E m+ is monomorphic Since the target group is in the Cartan-Eilenberg Ẽ-term converging to Ext Γ(m+(E m+, we have the composition η : Ext 0 G(m+(B m+ Ext Γ(m+(Em+ δ 0 Ext w w Γ(m+ So it is sufficient to show that η is monomorphic Since B m+ is in Ext 0 Γ(m+(N, we have the following diagram Ext 0 Γ(m+(M Ext 0 Γ(m+(N Ext Γ(m+(N v Ext Γ(m+ Ext 0 G(m+(B m+ η Ext Γ(m+ The right equality holds because Ext Γ(m+(M 0 = 0, and the top row is exact Since Ext 0 Γ(m+(M is the v A(m-module generated by vi /ip the map η is monomorphic as desired The Poincaré series of W m+ is given by g(w m+ = g(e m+ + g(b m+ = g(v m+ + g(um+ + g(b m+ = g m+ (t x x + x pj + x pj+ ( y pj j 0 x pj j 0 ( x pj+ ( x pj = g m+ (t x x + x pj+ = g m+ (t x pj x pj+ x pj j 0 j 0 = g(ext Γ(m+ by [Rav0, Theorem 37] x ( g Ext 0 G(m+(W m+ = x

9 ON β-elements IN THE ADAMS-NOVIKOV SPECTRAL SEQUENCE 9 This means that W m+ is weak injective by [Rav0, Theorem 6] 3 Basic methods for finding comodule primitives From now on, all Ext groups are understood to be over G(m + Definition 3 [Rav04, Definition 78] A G(m + -comodule M is called j-free if the comodule tensor product T (j m A(m+ M is weak injective, ie, Ext n (A(m +, T (j m A(m+ M = 0 for n > 0 The elements of Ext 0 are called j-primitives We will often abbreviate Ext(A(m +, N by Ext(N for short We will see in Proposition 33 that it is enough to consider a certain subgroup L j (M of M to detect elements of Ext 0 (T (j m M Given a right G(m + -comodule M and the structure map ψ M : M G(m + M, define the Quillen operation r i : M M (i 0 on z M by ψ M (z = i r i(z t i In this paper all comodules are right comodules In most cases the structure map is determined by the right unit formula Definition 3 The group L j (M Denote the subgroup n p j ker r n of M by L j (M By definition, we have a sequence of inclusions and L 0 (M = Ext 0 (M L 0 (M L (M L j (M The following result allows us to identify j-primitives with L j (M Proposition 33 [Rav0, Lemma ] Identification of the j-primitives with L j (m For a G(m + -comodule M, the map (c ψ M : L j (M Ext 0 (T (j w m M is an isomorphism between A(m + -modules, where c is the conjugation map When we detect elements of L j (M, it is enough to consider elements killed by r p j (j 0, as one sees by the following proposition Proposition 34 A property of Quillen operations If the Quillen operation r p j on a G(m + -comodule M is trivial, then all operations r n for p j n < p j+ are trivial Proof Since r i r j = ( i+j i ri+j [Nak, Lemma 3] we have a relation r n p j r p j = ( n p rn Observing that the congruence ( n j p s mod (p for j sp j n < (s + p j, ( n p is invertible in j Z(p whenever p j n < p j+, and the result follows In the following sections we will determine the structure of L 0 (B m+ in Proposition 4 and 44 and L (B m+ in Proposition 5 and 54 in all dimensions, and L j (B m+ (j > in Theorem 6 below dimension v pj + /v pj Then we need a method for checking whether all j-primitives (j > are listed or not The following lemma gives an explicit criterion the j-freeness of a comodule M

10 0 HIROFUMI NAKAI AND DOUGLAS C RAVENEL Lemma 35 A Poincaré series characterization of j-free comodules For a graded torsion connective G(m+-comodule M of finite type, we have an inequality (36 g(m( x pj g(l j (M where x = t v with equality holding iff M is j-free Proof Let I A(m + be the maximal ideal We have the inequality g(t (j m M g(ext 0 (T (j m M g(g(m + /I by [Rav04] Theorem 734, where the equality holds iff M is a weak injective Observe that and g(t (j m M = g(m xp j x, g(g(m + /I = x g(ext 0 (T (j m M = g(l j (M Lemma 37 A Poincaré series formula for the first Ext group For a graded torsion connective G(m + -comodule M of finite type, suppose g(l j (M x pj g(m ct d mod t d+ Then the first nontrivial element in Ext (T (j m M occurs in dimension d, and the order of the group G = Ext,d (T (j m M is p c Proof Since the inequality of (36 is an equality below dimension d, M is j-free in that range, so Ext (T (j m M vanishes below dimension d Each element x G is represented by a short exact sequence of the form 0 T (j m M If x has order p i, then we get a diagram 0 T (j m M M M Σ d A(m + 0 Σ d A(m T (j m M M Σ d A(m + /(p i 0 Since G is a finite abelian p-group, it is a direct sum of cyclic groups We can do the above for each of its generators and assemble them into an extension 0 T (j m M M Σ d G Z(p A(m + 0 with Ext 0 G(m+(M = L j (M through dimension d and Ext,d G(m+ (M = 0, so M is weak injective through dimension d

11 ON β-elements IN THE ADAMS-NOVIKOV SPECTRAL SEQUENCE If G = p b, then we have g(m = g(t (j m M + g(σ d G Z(p A(m + ( x pj = g(m + bt d g m+ (t x Since M is weak injective through dimension d, we have ( g(m g Ext 0 G(m+(M x mod t d+ g (L j(m x ( x pj g(m x + ct d so b = c 4 0-primitives in B m+ In this section we determine the structure of Ext 0 (B m+, ie, the primitives in B m+ in the usual sense We treat the cases m > 0 and m = 0 separately The latter is more complicated because v is not invariant over Γ( Recall that the G(m + -comodule structure of B m+ is given by the right unit map η R Lemma 4 An approximation of the right unit The right unit map η R : A(m + G(m + on the Hazewinkel generators are expressed by where ω = p m η R ( v = v + p t, η R ( v v + v t p vpω t mod (p Proof These directly follow from [MRW] ( and (3 For a given integer n, denote the exponent of a prime p in the factorization of n by ν p (n as usual In particular, ν p (0 = When the integer is a binomial coefficient ( n k, we will write νp ( n k instead of νp (( n k Let ĥj be the -dimensional cohomology class of t pj Proposition 4 Structure of Ext 0 (B m+ for m > 0 For m > 0, Ext 0 (B m+ is the A(m-module generated by { p k v s β ip k /t : i > 0, s 0, k 0, 0 < t pk and ν p (i ν p (s The first nontrivial element in Ext (B m+ is ĥ 0 β Ext,(p+(pω (B m+

12 HIROFUMI NAKAI AND DOUGLAS C RAVENEL Proof We may put s = ap l and i = bp l with p / b and a 0 Observe that ψ ( v apl v bpl+k bp l+ v t l = vap ( v pk + vpk pk+ t v pk+ ω t k bpl bp l+ v t since p / b l = vap v bpl+k bp l+ v t since t p k and so the exhibited elements are invariant On the other hand, we have nontrivial Quillen operations r (p k v s β ip k /t = vs v ip k p k v t pω and r p k+(p k v s β ip k /t = vs v p k (i pv t pk + s i vs v ipk v t if ν p (s < ν p (i + if t > p k, where the missing terms in the second expression involve lower powers of v in the numerator or smaller powers of v in the denominator This means each element p k v s β ip k /t with ν p(s < ν p (i supports a nontrivial r, the targets of which are linearly independent Similarly, each such monomial with t > p k supports a nontrivial r p k+ It follows that no linear combination of such elements is invariant, so Ext 0 is as stated For the second statement, note that ĥ0 and β are the first nontrivial elements in Ext and Ext 0 (B m+ respectively, so if their product is nontrivial, the claim follows It is nontrivial because there is no x B m+ with r (x = β We now turn to the case m = 0 Lemma 43 Right unit in G( The right unit η R : A( G( on the chromatic fraction is ipv t ( η R ipv t = ( t + k ( t k k ip k 0 k v t+k Note that this sum is finite because a chromatic fraction is nontrivial only when its denominator is divisible by p Proof Recall the expansion (x + y t = (x + y t = x t ( + y/x t = x ( t y t k k x k k 0 ( t + k ( y k = k 0 k x k+t and the formula η R (v t = (v + pt t by Lemma 4 Proposition 44 Structure of Ext 0 (B For m = 0, Ext 0 (B is the Z (p - module generated by { p k β ip k /t : i > 0, k 0, 0 < t pk and ν p (i ν p (t

13 ON β-elements IN THE ADAMS-NOVIKOV SPECTRAL SEQUENCE 3 The first nontrivial element in Ext (B is h 0 β Ext,(p (B m+ Proof When i and t are as stated, we may set t = ap l and i = bp l with p / b and a > 0 Observe that ( v bpl+k η R = (v pk bp l+ v apl + tpk+ bp l vpk v pk+ t pk ( ap l + n ( t n n 0 n bp l+ n v apl +n For n > 0, the binomial coefficient is divisible by p l+ n by Lemma A3 below, so the expression simplifies to η R ( v bpl+k bp l+ v apl k = (vp + tpk+ vpk v pk+ t pk bpl bp l+ v apl and p k β ip k /t is invariant by an argument similar to that of Lemma 4 On the other hand if either of the conditions on i and t fails, we have nontrivial Quillen operations or r p k+ r (p k β ip k /t ( p k β ip k /t k = vip p k v t p k = v(i p pv t pk t k i vip v t+ if ν p (i > ν p (t if t > p k The rest of the argument, inclduing the identifation of the first nontrivial element in Ext (B, is the same as in the case m > 0 5 -primitives in B m+ In this section we determine the structure of L (B m+, which includes all elements of Ext 0 (B m+ determined in the previous section By observing that r ( v β p = β p and r p j ( v β p = 0 for j, the first element of the quotient L (B m+ /L 0 (B m+ is v β p for m > 0 In general, we have Proposition 5 Structure of L (B m+ for m > 0 For m > 0, L (B m+ is isomorphic to the A(m-module generated by p k v s β ip k /t, where i > 0, s 0, k 0 and 0 < t p k, and the integers i and s satisfy the following condition: there is a non-negative integer n such that s 0,, p mod (p n+ and ν p (i < n + p Note that the description of L (B m+ differs from that of L 0 (B m+ given in Proposition 4 only in the restriction on i and s In that case it was ν p (i ν p (s If ν p (s = n + (ie, s 0 mod (p n+, then an integer i satisfying ν p (i n + also satisfies ν p (i < n + p Hence we have L 0 (B m+ L (B m+ as desired

14 4 HIROFUMI NAKAI AND DOUGLAS C RAVENEL Proof In Proposition 4 we have already seen that p k β ip k /t is invariant iff 0 < t p k If follows that r p l(p k v ( s β s ip k /p k = r p l( vs p k β ip k /p = v ipk k ppl p l v s pl ipv pk Since we are dealing with -primitives, we can ignore the case l = 0 For l =, this is clearly trivial if s < p When s p, choose an integer n such that p n ( s p By Lemma A4 this means n = 0 unless s is p-adically close to an integer ranging from 0 to p Then r p is trivial if ν p (i < n + p We can show that all Quillen operations r p l for l > are trivial under the same condition since ( ( ( s sp ν p p p ν p (p pl p l which follows from qν p ( p pl ( s p l = p l + + α(s p l α(s and ( ( ( ] sp s q [ν p (p pl l ν p p p p by Lemma A = p l p + α(s p l α(s p α(p l p + α(s p l α(s p 0 Note also that the condition on i and s in Proposition 5 is automatically satisfied whenever i < p p, which means that we may set n = 0 Since ( s r p ( v s = p p v s p p and p p kills all of B m+ below the dimension of β pp /p p, v is effectively invariant in this range, making B m+ an A(m + -module Corollary 5 Poincaré series for L (B m+ For m > 0, the Poincaré series for L (B m+ below dimension p p v is (53 g m+ (t k 0 and in the same range we have L (B m+ = A(m + x pk+ x pk, x pk { p k β ip k /t : i > 0, k 0 and 0 < t pk Proof As is explained in the above, we may consider L (B m+ as an A(m + - module in that range To determine the Poincaré series g(l (B m+, decompose L (B m+ into the following two direct summands: ( S 0 = A(m + /I { β i : i > 0 { ( S k = A(m + /I p k β ip k /t : i > 0 and pk < t p k for k > 0

15 ON β-elements IN THE ADAMS-NOVIKOV SPECTRAL SEQUENCE 5 The Poincaré series for these sets are given by g(s 0 = g m+ (t ( y n 0 and g(s k = g m+ (t ( y n>0 which gives y = g m+ (t n 0(y pk y pk x p n x pn y pk ( y pk p k y x pn+k x pn+k x pn+k x pn+k g(l (B m+ g m+ (t = n 0(y = n 0(y x pn x pn x pn x pn + 0<k n (y pk y pk + n>0(y pn y x pn x pn x pn x pn which is equal to (53 x = (y + (y pn x n>0 = n 0 x pn (y pn x pn x pn x pn Now we turn to the case m = 0, for which we make use of Lemma 43 again Observing that r (β p = β p/ and r p j (β p = 0 for j, the first element of the quotient L (B m+ /L 0 (B m+ is β p In general, we have Proposition 54 Structure of L (B For m = 0, L (B is isomorphic to the Z (p -module generated by p k β ip k /t, where k 0, i > 0 and 0 < t pk satisfying the following condition: there is a non-negative integer n such that t = 0,,, p mod (p n+ and p p+n / i Proof We have ( v ipk ψ ipv t = (v pk + tpk+ vpk v pk+ t ( t + r pk i r r 0 ( pt r ipv t+r in which there are terms v (i pk t pk+ pv t pk k, v(i p t pk pv t pk+ ( t + p l v ip k and ( p pl t pl p l ipv t+pl Since t p k, the first and the second are trivial, which gives ( ( r p l p k β ipk /t = ( p p l t + p l v ip k p l ipv t+pl for l 0

16 6 HIROFUMI NAKAI AND DOUGLAS C RAVENEL ( Choose an integer n such that p n t+p p, which occurs iff t = 0,,, p mod (p n+ by Lemma A4 Then r p is trivial if p p+n / i We can ( also observe that all the higher Quillen operations r l (l are trivial since ν p p p( ( t+p p ν p (p pl t+p l (see the proof of Proposition 5 p l Corollary 55 L (B as an A(-module For m = 0, we have L (B = A( {p k β ipk/t : i > 0, k 0 and 0 < t pk below dimension p p v The Poincaré series for L (B in this range is the same as (53 Applying Lemma 35 and 37 to the Poincaré series (53, we have the following result Corollary 56 -free range for B m+ For m 0, B m+ is -free below dimension p(p + v, and the first element in Ext (T ( m B m+ is β p/p ĥ Here we use the notation β p/p for its image under the map (c ψ Bm+ (cf (33 Proof By comparing g(b m+ and g(l (B m+ and using Lemma 37, we see that the first nontrivial element of Ext (T ( m B m+ occurs in the indicated dimension, where the group has order p The fact that β p/p ĥ is nontrivial in Ext follows by direct calculation 6 j-primitives in B m+ for j > In this section we determine the structure of L j (B m+ for j and m > 0 (See [Rav04] Lemma 73 for the m = 0 case The first element of the quotient L j (B m+ /L j (B m+ is β p j +/p j +, which has nontrivial Quillen operation r p j ( βp j +/p j + = β In general, we have Theorem 6 Structure of L j (B m+ in low dimensions for j > (i Below dimension p j+ v, L j (B m+ is the A(m + -module generated by { β i/t : 0 < t min(i, p j { βap j +b/t : p j < t p j, a > 0 and 0 b < p j (ii B m+ is j-free below dimension v pj+ v (iii The first element in Ext is the p-fold Massey product β +p j /p j, ĥ,j,, {{ ĥ,j p For the basic properties of Massey products, we refer the reader to [Rav86, A4] or [Rav04, A4]

17 ON β-elements IN THE ADAMS-NOVIKOV SPECTRAL SEQUENCE 7 Proof (i The listed elements are the only j-primitives below dimensions p j+ v by Proposition B3, and the first statement follows (ii To show that B m+ is j-free below the indicated dimension, we need to compute some Poincaré series This will be a lengthy calculation Decompose L j (B m+ into the following three direct summands: S 0, = A(m + { β i/t : 0 < t i < p j, S 0, = A(m + { β i/t : 0 < t p j i, S j = A(m + { βapj +b/t : p j < t p j, a > 0 and 0 b < p j We will always work below the dimension of β pj /pj, which is vpj+ v pj This means that in the description of S j above, the only relevant value of a is Observe that so S 0, = A(m + /I { v ipk p k v ipk l g(s 0, = g(a(m + /I g(s 0, g m+ (t = g m+ (t = = For S 0,, we have which is the quotient of by (x ip 0<i<p j k ( x pk ( (x pk pj k (x pk x pj xp x pk : 0 l < ip k, 0 < i < p j k ( k (x yip pk i y 0<i<p j k k x ipk k xp x pk k x pj x pk ( (x pk pj k x pk { v i S 0, = A(m + : 0 l < p j, i p j, ipv pj l { A(m + /I k>0 A(m + /I { v ipk p k v pj l v ipk p k v pj l : 0 l < p j, i > 0 : 0 l < p j, 0 < i < p j k,

18 8 HIROFUMI NAKAI AND DOUGLAS C RAVENEL Hence the Poincaré series of S 0, is ( ypj y pj g(s 0, = g(a(m + /I y (x pk i (x p k>0 i>0 0<i<p j k g(s 0, g m+ (t = (y pj x pk k>0 x pk = (y pj k>0 = (y pj k>j in our range of dimensions Adding these two gives g(s 0, S 0, g m+ (t We also observe that (y pj x pj + k i x pk ( (x pk pj k x pk x pk x pk x pk x pk + x pk 0<k j 0<k j x pj x pj x pk = g(s 0, + g(s 0, g m+ (t = (x pk x pj xp x pk = = k x pj x pk +(y pj x pj + 0<k j (x pk x pj + xp j x pk x pk x pk +(y pj x pj ( x pj (x pk x pk ( x pk ( x pk +x pj+ (y qpj y pj g(s j = g(a(m + /I xpj+ ( y qpj y = g m+ (t xpj+ x pj x pk x pj x pk x pj x pj x pk x p j + xpj x pj x p j + xpj x pj ( y qpj ( x pj x j xp x

19 ON β-elements IN THE ADAMS-NOVIKOV SPECTRAL SEQUENCE 9 Summing these three Poincaré series, we obtain g(s 0, S 0, S j g m+ (t = g(s 0, + g(s 0, + g(s j g m+ (t = = = = ( x pj (x pk x pk ( x pk ( x pk + xpj x p j x pj xpj+ ( y qpj ( x pj +x pj+ (y qpj y pj + x ( x pj (x pk x pk ( x pk ( x pk + xpj x p j x pj xpj+ (( y qpj ( x pj + (y qpj y pj ( x + x ( x pj (x pk x pk ( x pk ( x pk x p j + xpj x pj xpj+ ( x pj + y qpj x pj y pj x y qpj + x y pj + x ( x pj (x pk x pk ( x pk ( x pk On the other hand, Theorem 4 gives + xpj x p j x pj xpj+ ( x pj y qpj (x x pj y pj ( x + x g(b m+ g m+ (t 0<k j+ x pk x pk ( x pk ( x pk x pk x pk ( x pk ( x pk + xp j x pj ( x pj ( x pj + xp j+ x pj x pj+ below dimension x pj+ x pj, so g(b m+ ( x pj g m+ (t = (x pk x pk ( x pj ( x pk ( x pk + xpj+ ( y pj ( x pj x pj+ + xpj x p j x pj

20 0 HIROFUMI NAKAI AND DOUGLAS C RAVENEL This means g(s 0, S 0, S j g(b m+ ( x pj g m+ (t xpj+ ( x pj y qpj (x x pj y pj ( x = x xpj+ ( y pj ( x pj x pj+ xpj+ ( y qpj x y pj ( x xp x below dimension v pj (p+ xpj+ x ( y qpj = x j+( y pj x + x y pj x By Lemma 35, this means that B m+ is j-free in the range claimed and that the first nontrivial Ext has order p (iii To show that the generator of is Ext the element specified, we first show that the indicated Massey product is defined For j > and < k < p we claim d( β +kp j /kp j = β +p j /p j, ĥ,j,, {{ ĥ,j k This can be shown by induction on k and direct calculation as follows Let s = t p vpω t T (j m G(m + It follows that w = v v s is invariant Note that its p j th power does not lie in T (j m Then we have η R ( β+kp j /kp j = η R ( v kpj = 0<l k = 0<l k = 0<l k w pv kpj ( kp j v lp j w lp j pv lpj ( k v lp j l ( k l w pv lpj s (k lpj s (k lpj β +lp j /lpj s(k lpj = β +p j /p j, ĥ,j,, {{ ĥ,j k This means that our p-fold Massey product is defined

21 ON β-elements IN THE ADAMS-NOVIKOV SPECTRAL SEQUENCE We claim the first element in Ext is represented by ( p β p l +lp j /lpj s(p lpj 0<l<p = ( p β +lp j /lp ( t j pj 0<l<p = 0<l<p p p l ( p l β +lp j /lp j t pj (p l = β +qp j /qp j t pj + vpj p l (pω t pj The only element in B m+ in this dimension is β +pj /pj, which is primitive, so this element in Ext is notrivial 7 Higher Ext groups for j = In this section we exhibit some calculations of Ext s (T (j m B m+ for s > 0 Recall the small descent spectral sequence, constructed in [Rav0, Theorem 7], which converges to Ext(T (j m B m+ with E,s = E(ĥj P ( b j Ext(T (j+ m B m+ with ĥj E,0 and b j E,0, and d r : Er s,t Er s+r,t r+ In particular, d is induced by the action of r p j on B m+ for s even and r qp j for s odd The case m = 0 has already been treated in [Rav04, Chapter 7], so we may assume that m > 0 We examine the simplest case, j = Recall that B m+ is -free below dimension v p + /v p and Ext0 (T ( m B m+ is the A(m + -module generated by (7 { β i/t : 0 < t min(i, p { βp /t : p < t p by Theorem 6 Then the spectral sequence collapses from E We can compute d on elements (7 using Proposition B: The action of r p on Ext 0 (T ( m B m+ is given by r p ( β i/e = β i /e and r p ( βpi/e = 0, and the action of r qp is obtained by composing r p up to unit scalar In order to understand the behavior of d, the following picture for p = 3 may be helpful β β β 3 β3 [^ [ [^ r3 r [ [^ 3 r [ 3 (7 β / β 3/ β3/ β4/ [^ [ [^ r3 [ r 3 [ [^ r 3 β 3/3 β 3/3 β4/3 β5/3

22 HIROFUMI NAKAI AND DOUGLAS C RAVENEL Here each arrow represents the action of the Quillen operation r 3 up to unit scalar For a general prime p, the analogous picture would show a directed graph with p components, two of which have p vertices, and in which the arrow shows the action of the Quillen operation r p up to unit scalar Each component corresponds to an A(m + -summand of the E -term, with the caveat that p β p/e = β p/e and v β i/e = β i/e Notice that the entire configuration is vp -periodic Corresponding to the diagonal containing β in (7, the subgroup of E generated by { β, β /, β 3/3 E(ĥ, P ( b, reduces on passage to E to simply { β Similarly, the subset { β, β 3/ E(ĥ, P ( b, reduces to { β, β 3/ĥ, P ( b,, where β 3/ĥ, = ĥ,, ĥ,, β and ĥ, ( β 3/ĥ, = ĥ, ĥ,, ĥ,, β = ĥ,, ĥ,, ĥ, β = b, β These observations give us the following result Proposition 73 Structure of Ext(T ( m B m+ In dimensions less than v p + /v p (, Ext(T m B m+ is a free module over A(m + /I with basis { β pi+s ; β pi+p/s ; β p /l { β+pi, β p+pi ; β p /k P ( b, ĥ, { β pi+p/t ; β pi+r/p ; β p /l where 0 i < p, k p p +, p p + l p, s p, t p and p u p, subject to the caveat that v βp/e = β p/e and p β p/e = β p/e In particular Ext 0 (T ( m B m+ has basis { β +pi,, β p+pi; β p+pi/p,, β p+pi/ ; β p /p,, β p / Note that for m > 0, this range of dimensions exceeds p v 3 Appendix A Some results on binomial coefficients Fix a prime number p Definition A α(n, the sum of the p-adic digits of n For a nonnegative integer n, α(n denotes sum of the digits in the p-adic expansion of n, ie, for n = i 0 a ip i with 0 a i p, we define α(n = i 0 a i As before, let ν p (n denote the p-adic valuation of n, ie, the exponent that makes n a p-local unit multiple of p νp(n When the integer is a binomial coefficient ( i ( j, we will write i ( (i νp j instead of νp j Then we have,

23 ON β-elements IN THE ADAMS-NOVIKOV SPECTRAL SEQUENCE 3 Lemma A p-adic valuation of a binomial coefficient ( n qν p = α(k + α(n k α(n k where q = p In particular, qν p ( n p j = + α(n p j α(n Proof Recall that qν p (n! = n α(n, and observe that ( ( n n! qν p = qν p k (n k!k! = q (ν p (n! ν p ((n k! ν p (k! = n α(n (n k + α(n k k + α(k = α(n + α(n k + α(k Using this lemma we can determine the number how many times a binomial coefficient is divisible by a prime p For example, we have Lemma A3 Divisibility of a binomial coefficient Assume that p / a and 0 < n l Then the binomial coefficient ( ap l +n n is divisible by p l+ n Proof Since a 0 mod (p, we have α(a = α(a Let m = ν p (n and n = n p m Then α(n = α(n, and we have ( ap l + n qν p n ( ap l + n p m = qν p n p m = α(n p m + α(ap l α(ap l + n p m = α(n + α(a + ql α(ap l m + n qm = α(n + α(a + ql α(a α(n qm = q(l m q(l + n We consider this type of binomial coefficients in Proposition 44 The other types we need are the followings: Lemma A4 Divisibility of another binomial coefficient Assume that p is a prime and that a positive integer s is expressed as s = s p l + s 0 > 0 with 0 s 0 < p l Then we have ν p ( s p l = νp (s In particular, we have p n ( s p l iff s 0,,, p l mod (p n+l Proof Observe that qν p ( s p l = α(p l + α(s p l α(s = + α((s p l + s 0 α(s p l + s 0 = α( + α(s α(s = qν p (s

24 4 HIROFUMI NAKAI AND DOUGLAS C RAVENEL This implies that ν p ( s p l = n iff s s0 mod (p n+l In Appendix B it is required to know how many times the binomial coefficient ( i p j is divisible by p For 0 < i < p j it is clear that ( i p j = 0 For i p j, the number ν p ( i p j can be determined explicitly in the following results Proposition A5 A third divisibility statement For i p j, define nonnegative integers i 0 and i by (A6 i = i p j + i 0 ( i > 0 and 0 i 0 < p j Then we have ( ( i p j is divisible by p iff i0 0; ( More generally, ( i p j is divisible by p j k (0 k < j iff (A7 ν p (i 0 k + ν p (i or equivalently i 0 0 and p k+νp(i / i 0 In particular, the inequality (A7 is automatically satisfied if ν p (i j k Proof Observe that ( ( i i ν p p j = ν p (p j + ν p p j ν p (i { (j + νp (i = (j + ν p (i if i 0 = 0 ν p (i 0 if i 0 0 { 0 if i0 = 0 = j + ν p (i ν p (i 0 if i 0 0 by Lemma A4 If i 0 0, then we have j + ν p (i ν p (i 0 > 0 since ν p (i 0 j, and so the binomial coefficient is divisible by p Since i 0 = 0 is equivalent to p j i, the statement ( follows The condition p j k ( ( i p j is equivalent to the inequality i νp p j k, j and if we suppose that j k > 0 then this inequality gives (A7 Note that (A7 is always satisfied if ν p (i j k since ν p (i 0 j by definition The following is the obvious translation of Proposition A5 Corollary A8 A fourth divisibility statement Let i 0 and i be as in (A6 and assume that p j < i p j +m Then, we have p j k ( i p for 0 k < j j iff ν p (i 0 k + ν p (i with 0 ν p (i m Proof The given range p j < i p j +m means that 0 ν p (i m and the result follows from Proposition A5

25 ON β-elements IN THE ADAMS-NOVIKOV SPECTRAL SEQUENCE 5 Appendix B Quillen operations on β-elements In this section we discuss the action of the Quillen operations r p j the β-elements for j > 0 on First we consider the following easy cases Proposition B Primitive β-elements For i > 0, the elements β i/t are primitive if 0 < t p νp(i, ie, it satisfies r l ( β i/t = 0 for all l 0 Proof Set ν p (i = n and i = i p n By direct calculations we have ( v i η R pv t = ( vpn + v pn t pn+ v pn+ ω pv t t pn i = vi pv t For the other cases, the Quillen operation r p j is computed as follows: Proposition B Quillen operations on β-elements When j > 0, we have ( i r p j ( β i/t = β p j i p j /t p for t < j pj + p m+ Proof First assume that m > 0 Observe that ( v η R ( β i/t i = η R ipv t = ( ( i l = ( k l k = 0 k l i 0 k l i ( v + v t p vpω t i ipv t v i l ( ( i l ( k l k ( v t p l k ( v pω t k ipv t v i l t p(l k+k (i lpv t l+k pωk Since r p j ( β i/t is the coefficient of t pj in the above, we need to consider the terms satisfying p(l k + k = p j Note that k must be divisible by p and that we may set k = pn Thus we have Now let p j = p(l pn + pn l(n = l = p j + qn where q = p and g(n = t l + k pωk = t p j qn + pn p m+ n = t p j n(p m+ Then we have r p j ( β i/t = ( ( i l(n ( pn l(n np 0 n p j v i l(n (i l(npv g(n

26 6 HIROFUMI NAKAI AND DOUGLAS C RAVENEL Given our assumption about t, the only value of n satisfying g(n > 0 is n = 0, which gives ( i r p j ( β i/t = v i pj p j (i p j pv t pj The proof for m = 0 is more complicated Observe that ψ(β i/t = ( ( ( i l t + r ( p k+r r v i l t p(l k+k+r l k r (i lpv t+r l+k pk, 0 k l i r 0 which shows that r p j (β i/t is equal to 0 n p j 0 r np ( ( ( i l(n, r t + r p ( np r v i l(n,r l(n, r np r r (np rpv g(n,r where l(n, r = p j + nq r and g(n, r = t p j n(p + r(p + If p r (np r for a positive r, then we may put r = sp and n p sp + s for a positive s and the exponent of v is not positive since, g(n, r t p j (p sp + s(p + sp(p + = t p j (p + (p sp p sp s t p j (p + (p p p p t p j (p Thus, the nontrivial term arises only when r = 0 We can see that it is also required that n = 0 by the same reason as the m > 0 case, and the result follows To know the condition of triviality of r p j in Proposition B, we need the results on the p-adic valuation of binomial coefficients obtained in Appendix A In particular, we have Proposition B3 Some trivial actions of Quillen operations Assume that p j < i p j+ and t < p j + p m+ Then we have the following trivial Quillen operations: ( r p l( β i/t (l j for 0 < t min(i, pj ; ( r p l( β ap j +b/t (l j for p j < t p j and 0 b < p j Proof We will show the following Quillen operations on p k β i/t are trivial: (a r p l (l j for 0 < t min(i, p j and k 0; (b r p l (l j for p j < t p j, i = ap j +bp k with p / a, p / b and 0 k < j ; (c r p l (l 0 for p j < t p j, i = ap j with 0 < a p and j = k For the case (, note that ( i v i p j r p j (p k β i/t = p j p j k v t pj by Proposition B, which is clearly trivial when 0 < t p j ( p l Even if p j < t i, it is trivial when the binomial coefficient ( i p is divisible by p j k, j or equivalently when the inequality (A7 holds

27 ON β-elements IN THE ADAMS-NOVIKOV SPECTRAL SEQUENCE 7 When 0 < k < j, by the assumption we have p j < i p j + i 0 p j+ (where ν p (i 0 < j by definition and ν p (i Note that if k > 0 and p k / i then p k β i/t itself is trivial and that we may assume that ν p(i k These observations suggest that the only case satisfying the inequality (A7 is (ν p (i, ν p (i 0 = (, k, which gives the case (b When j = k, the Quillen operation r p j (p j β i/t is clearly trivial and pj β i/t is nontrivial only if p j i, which gives the case (c For the case (b and (c, observe that the Quillen operation r p j+(p k β i/t is a unit scalar multiple of β i p j /t p j and pk β i/t is not in L j(b m+, which means that the condition t p j is required Conbining (b and (c gives the case ( Note that no linear combination of β-elements can be killed by r p j since the r p j -image has different exponents of v or v if β i /t β i /t References [MRW] HRMiller, DCRavenel and WSWilson, Periodic phenomena in the Adams-Novikov spectral sequence Ann Math (, 06:469 56, 977 [Nak] H Nakai An algebraic generalization of Image J, To appear in Homology, Homotopy and Applications [NR] HNakai and D C Ravenel The method of infinite descent in stable homotopy theory II in preparation [Rav86] D C Ravenel Complex Cobordism and Stable Homotopy Groups of Spheres Academic Press, New York, 986 [Rav0] D C Ravenel The method of infinite descent in stable homotopy theory I In D M Davis, editor, Recent Progress in Homotopy Theory, volume 93 of Contemporary Mathematics, pages 5 84, Providence, Rhode Island, 00 American Mathematical Society [Rav04] D C Ravenel Complex Cobordism and Stable Homotopy Groups of Spheres, Second Edition American Mathematical Society, Providence, 004 Available online at Department of Mathematics,Musashi Institute of Technology,Tokyo , Japan address: nakai@mansmusashi-techacjp Department of Mathematics,University of Rochester, Rochester, New York 467 address: douglasravenel@rochesteredu